## 42.26 The key formula

Let $(S, \delta )$ be as in Situation 42.7.1. Let $X$ be locally of finite type over $S$. Assume $X$ is integral and $\dim _\delta (X) = n$. Let $\mathcal{L}$ and $\mathcal{N}$ be invertible sheaves on $X$. Let $s$ be a nonzero meromorphic section of $\mathcal{L}$ and let $t$ be a nonzero meromorphic section of $\mathcal{N}$. Let $Z_ i \subset X$, $i \in I$ be a locally finite set of irreducible closed subsets of codimension $1$ with the following property: If $Z \not\in \{ Z_ i\}$ with generic point $\xi$, then $s$ is a generator for $\mathcal{L}_\xi$ and $t$ is a generator for $\mathcal{N}_\xi$. Such a set exists by Divisors, Lemma 31.27.2. Then

$\text{div}_\mathcal {L}(s) = \sum \text{ord}_{Z_ i, \mathcal{L}}(s) [Z_ i]$

and similarly

$\text{div}_\mathcal {N}(t) = \sum \text{ord}_{Z_ i, \mathcal{N}}(t) [Z_ i]$

Unwinding the definitions more, we pick for each $i$ generators $s_ i \in \mathcal{L}_{\xi _ i}$ and $t_ i \in \mathcal{N}_{\xi _ i}$ where $\xi _ i$ is the generic point of $Z_ i$. Then we can write

$s = f_ i s_ i \quad \text{and}\quad t = g_ i t_ i$

Set $B_ i = \mathcal{O}_{X, \xi _ i}$. Then by definition

$\text{ord}_{Z_ i, \mathcal{L}}(s) = \text{ord}_{B_ i}(f_ i) \quad \text{and}\quad \text{ord}_{Z_ i, \mathcal{N}}(t) = \text{ord}_{B_ i}(g_ i)$

Since $t_ i$ is a generator of $\mathcal{N}_{\xi _ i}$ we see that its image in the fibre $\mathcal{N}_{\xi _ i} \otimes \kappa (\xi _ i)$ is a nonzero meromorphic section of $\mathcal{N}|_{Z_ i}$. We will denote this image $t_ i|_{Z_ i}$. From our definitions it follows that

$c_1(\mathcal{N}) \cap \text{div}_\mathcal {L}(s) = \sum \text{ord}_{B_ i}(f_ i) (Z_ i \to X)_*\text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i})$

and similarly

$c_1(\mathcal{L}) \cap \text{div}_\mathcal {N}(t) = \sum \text{ord}_{B_ i}(g_ i) (Z_ i \to X)_*\text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i})$

in $\mathop{\mathrm{CH}}\nolimits _{n - 2}(X)$. We are going to find a rational equivalence between these two cycles. To do this we consider the tame symbol

$\partial _{B_ i}(f_ i, g_ i) \in \kappa (\xi _ i)^*$

see Section 42.5.

Lemma 42.26.1 (Key formula). In the situation above the cycle

$\sum (Z_ i \to X)_*\left( \text{ord}_{B_ i}(f_ i) \text{div}_{\mathcal{N}|_{Z_ i}}(t_ i|_{Z_ i}) - \text{ord}_{B_ i}(g_ i) \text{div}_{\mathcal{L}|_{Z_ i}}(s_ i|_{Z_ i}) \right)$

is equal to the cycle

$\sum (Z_ i \to X)_*\text{div}(\partial _{B_ i}(f_ i, g_ i))$

Proof. First, let us examine what happens if we replace $s_ i$ by $us_ i$ for some unit $u$ in $B_ i$. Then $f_ i$ gets replaced by $u^{-1} f_ i$. Thus the first part of the first expression of the lemma is unchanged and in the second part we add

$-\text{ord}_{B_ i}(g_ i)\text{div}(u|_{Z_ i})$

(where $u|_{Z_ i}$ is the image of $u$ in the residue field) by Divisors, Lemma 31.27.3 and in the second expression we add

$\text{div}(\partial _{B_ i}(u^{-1}, g_ i))$

by bi-linearity of the tame symbol. These terms agree by property (6) of the tame symbol.

Let $Z \subset X$ be an irreducible closed with $\dim _\delta (Z) = n - 2$. To show that the coefficients of $Z$ of the two cycles of the lemma is the same, we may do a replacement $s_ i \mapsto us_ i$ as in the previous paragraph. In exactly the same way one shows that we may do a replacement $t_ i \mapsto vt_ i$ for some unit $v$ of $B_ i$.

Since we are proving the equality of cycles we may argue one coefficient at a time. Thus we choose an irreducible closed $Z \subset X$ with $\dim _\delta (Z) = n - 2$ and compare coefficients. Let $\xi \in Z$ be the generic point and set $A = \mathcal{O}_{X, \xi }$. This is a Noetherian local domain of dimension $2$. Choose generators $\sigma$ and $\tau$ for $\mathcal{L}_\xi$ and $\mathcal{N}_\xi$. After shrinking $X$, we may and do assume $\sigma$ and $\tau$ define trivializations of the invertible sheaves $\mathcal{L}$ and $\mathcal{N}$ over all of $X$. Because $Z_ i$ is locally finite after shrinking $X$ we may assume $Z \subset Z_ i$ for all $i \in I$ and that $I$ is finite. Then $\xi _ i$ corresponds to a prime $\mathfrak q_ i \subset A$ of height $1$. We may write $s_ i = a_ i \sigma$ and $t_ i = b_ i \tau$ for some $a_ i$ and $b_ i$ units in $A_{\mathfrak q_ i}$. By the remarks above, it suffices to prove the lemma when $a_ i = b_ i = 1$ for all $i$.

Assume $a_ i = b_ i = 1$ for all $i$. Then the first expression of the lemma is zero, because we choose $\sigma$ and $\tau$ to be trivializing sections. Write $s = f\sigma$ and $t = g \tau$ with $f$ and $g$ in the fraction field of $A$. By the previous paragraph we have reduced to the case $f_ i = f$ and $g_ i = g$ for all $i$. Moreover, for a height $1$ prime $\mathfrak q$ of $A$ which is not in $\{ \mathfrak q_ i\}$ we have that both $f$ and $g$ are units in $A_\mathfrak q$ (by our choice of the family $\{ Z_ i\}$ in the discussion preceding the lemma). Thus the coefficient of $Z$ in the second expression of the lemma is

$\sum \nolimits _ i \text{ord}_{A/\mathfrak q_ i}(\partial _{B_ i}(f, g))$

which is zero by the key Lemma 42.6.3. $\square$

## Comments (2)

Comment #2981 by Xia on

There seems to be some typos in the first displayed formula in the proof of Lemma 41.26.1 and the bracket after it.

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